By definition, we say that a (sub)sequence converges if its limit exists (and is finite).
By definition, if the limit exists, it is unique. You may prove this for yourself.

For example, the sequence {-1, 1, -1, 1, -1, 1, ....} has two converging subsequences {1, 1, 1, 1, ...} and {-1, -1, -1, -1, ...} (actually, there are infinitely many, as long as it ends in just 1's or just -1's) which converge to 1 and -1 respectively. The sequence itself has no limit though.

there is a theorem i guess, which states that if {an} converges then every subsequence of it converges and that to the same nr as {an}. Hence if we can find at least two subsequences of a sequence {an} that converge do different nrs, that is have different limits, then the sequence {an} does not converge!

Specifically, you can do this: suppose the subsequence of {an} has subsequence {an}i which converges to P and subsequence {an}j which converges to Q.

Assume that {an} converges to L, and take [itex]\epsilon[/itex]= (1/2)|P- Q|. For any N, there will be n1> N such that an1, in the first subsequence, is arbitratily close to P and n2> N such that an2, in the second subsequence is arbitrarily close to Q. If they are not within [itex]2\epsilon[/itex] of each other, they cannot both be within [itex]\epsilon[/itex] of L, a contradiction.

We need to show that [tex]\forall \epsilon>0[/tex] [tex] \exists I[/tex] such that [tex]i\geq I[/tex] implies [tex]|a_{n_i}-L|<\epsilon[/tex]. But if we choose [tex]I[/tex] such that [tex]i\geq I[/tex] implies that [tex]n_i\geq N[/tex], then [tex]i\geq I[/tex] implies [tex]|a_{n_i}-L|<\epsilon[/tex]. The existence of such an [tex]I[/tex] is guaranteed by the definition of a subsequence (something to check).

So what this proof says is that if a sequence is convergent, then all it's subsequences converge to the same limit. Now take the contra-positive of this statement and compare it with your question.